Amazon·Data Scientist·Technical Phone Screen·Intermediate

Intermediate

May 2026

Summary

Amazon data scientist interview that leaned heavily into streaming algorithms and probability theory. The coding portion felt more like a stats exam than a typical LC grind, which I wasn't fully prepared for.

Questions Asked (2)

Implement reservoir sampling for an unbounded stream of user IDs to maintain a uniform random sample of k users in memory. Walk through the time and space complexity, and explain why every element ends up with exactly k/n probability of being selected.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

This one took me a minute to get grounded.

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AI HintsAI Generated

Suggested Approach

Start by explaining the reservoir sampling algorithm: initialize a reservoir with the first k elements, then for each subsequent element i (1-indexed), generate a random integer j between 1 and i; if j <= k, replace the j-th element in the reservoir with the current element. Then analyze time and space complexity (O(n) time, O(k) space) and prove the uniform probability k/n using induction or by computing the probability that any element is included.

Pro tip: Mention that reservoir sampling is ideal for streaming data where the total size is unknown, and highlight its use in Amazon's real-time analytics pipelines for unbiased sampling. Also, note that the algorithm can be adapted for weighted sampling if needed.

1. Describe the algorithm

Clearly outline the steps: fill reservoir with first k elements; for each subsequent element i, pick a random index j from 1 to i; if j <= k, replace reservoir[j] with element i.

2. Analyze complexity

State that time complexity is O(n) since each element is processed once, and space complexity is O(k) for the reservoir. Emphasize that it's a single-pass algorithm suitable for streams.

3. Prove uniform probability

Show that each element has probability k/n of being in the final reservoir. Use induction: for the first k elements, probability is 1 initially, but after processing all n, it becomes k/n. For element i > k, probability of being selected is k/i, and probability of surviving subsequent replacements is i/(i+1) * (i+1)/(i+2) * ... * (n-1)/n = i/n, so overall probability = (k/i)*(i/n) = k/n.

4. Discuss practical considerations

Mention handling of edge cases (e.g., n < k), and that the algorithm works even if n is unknown. Also, note that it's unbiased and can be extended to weighted reservoir sampling.

Key Points to Mention

Reservoir sampling maintains a uniform random sample of size k from a stream of unknown length.
Time complexity: O(n) for n elements; space complexity: O(k).
Probability proof: each element has probability k/n of being in the final sample.
The algorithm is single-pass and requires no knowledge of n in advance.
It is used in real-time data pipelines for unbiased sampling, e.g., at Amazon for user analytics.
Edge cases: when n < k, all elements are included; when k=1, it reduces to random selection.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.

Given an array of integers, return the indices of any duplicate values. You can only use a set and enumerate. The solution must run in O(n) time.

Algorithms & Data Structures

Author's notes

Easier than the first question.

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AI HintsAI Generated

Suggested Approach

Use a set to track seen numbers while iterating through the array with enumerate. When a number is already in the set, return its current index and the stored index of its first occurrence. This ensures O(n) time and O(n) space.

Pro tip: Clarify upfront that you'll return the first duplicate found and its previous index, and mention that using a set gives average O(1) lookups. This shows you consider edge cases and performance.

1. Clarify requirements

Confirm that any duplicate is acceptable, and decide whether to return the first duplicate encountered or all duplicates. Also confirm the return format (e.g., list of index pairs).

2. Initialize data structures

Create an empty set to store seen numbers and a dictionary to map numbers to their first index (or just store indices in the set if only the first duplicate is needed).

3. Iterate with enumerate

Loop through the array using enumerate to get both index and value. For each value, check if it's in the set.

4. Handle duplicates

If the value is already in the set, return the stored index and the current index. Otherwise, add the value to the set and store its index.

5. Return result

If no duplicates are found after the loop, return an empty list or a message indicating no duplicates.

Key Points to Mention

Time complexity: O(n) because each element is processed once with O(1) set operations.
Space complexity: O(n) in the worst case when all elements are unique.
Use of enumerate to efficiently get index and value in one pass.
Set provides average O(1) membership check, crucial for O(n) time.
Edge cases: empty array, no duplicates, multiple duplicates.
Return format: indices of any duplicate (e.g., [i, j] where arr[i] == arr[j]).

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.